Graphing Simple Rational Functions

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Question

Graphing Simple Rational Functions

Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.

y = (x + 3)/(x + 2)

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Draw the graph of the rational function. Y is equal to x minus 4 divided by X + 1, then write the equations for its asymptotes and identify its intercepts. Awesome. So it appears for this particular problem, we're ultimately trying to create a graph of this rational function of Y. That's our first answer we're trying to solve for. And then we're ultimately trying to figure out how to write the equations for the asymptotes, and we're also asked to identify the intercept points. Awesome. So once again, we're trying to solve for the, once again, the equations for the asymptotes, whether they're vertical and or horizontal asceopes, and we're also asked to identify the intercepts for both X and Y intercepts if they exist. So, With that said, now that we know what we're ultimately solving for, let us try to solve for vertical asymptotes first. So, Let us start with vertical asymptotes. So now that we know that we're solving for vertical asymptotes, as we should recall and note, vertical asymptotes occur where the denominator of the function will be equal to 0, assuming that the numerator does not also equal 0 at that specific point. So we need to set the denominator equal to 0, so meaning we need to take X + 1 is equal to 0, so that will mean when we solve for X, that X is going to be equal to -1. So that means that there will be a vertical asymptote at X equals 1. So let's note that at X equals -1, there will be a vertical asymptote. So let's put a box around that because that is one of our final answers. Hooray, we're on a roll here. We found one of our answers that we're trying to solve for. So now, our next step is we're trying to determine whether or not they are horizontal asymptotes. So as we should recall and note, in order to find horizontal asymptotes, we need to examine the degree of the numerator and the denominator. So the numerator, as we should recall, is a polynomial with a degree of one. And the denominator is a polynomial with a degree of one as well. So the denominator also has a degree of one. So when the degrees of the numerator and denominator are equal, that means that the horizontal asymptote is determined by the ratio of the leading coefficients. And in this particular case for our problem, the leading coefficient of the numerator is 1, because we have X, which is equivalent to saying there's like a 1, an invisible one out front of the X, so it's 1 X, so they're both 1 X or just X. So, so therefore, in this particular case, the leading coefficient of the denominator is also one because they're both 1 X or just X. So therefore, that will mean that the horizontal asymptote is, as we should recall, Y is equal to 1 divided by 1, which will mean that it's equal to 1. So, therefore, without a doubt, we can safely say that our horizontal acetode is Y is equal to 1, and that is for our horizontal asymptote. Hooray, we solved for our one of our final answers, since we're trying to figure out the equations for the asymptotes and then identify. It's so now we're trying to say that so once again we're trying to identify the intercepts and then also the asymptoms. So now, moving right along, so focusing our attention now on dominant terms. So in the function, and let us note that our original function states that Y is equal to X minus 4 divided by X plus 1, the dominant term is as X approaches positive infinity. That means, or, so it's once again, so in the function for our rational function of Y, the dominant term as X approaches positive infinity or as X approaches negative infinity is determined by the highest degree terms in. The numerator and denominator. So the dominant term will be X divided by X, which is equal to 1. So this confirms that the horizontal asymptote is indeed, Y is equal to 1. So now we can shift gears. And start to solve for our intercepts, which I'll just simply call I. So to find our intercepts, let us focus our attention on solving for the Y intercept first. So our Y intercept, in order to find our Y intercept, as we should recall, we need to set X is equal to 0 in our equation, so we need to say that Y is equal to 0, minus 4 divided by 0 + 1, which will give us an answer of -4. So that will mean that our Y intercept is at parenthesis negative, so it's gonna, once again, we found an answer of -4. So that will mean that our Y intercept will be at parentheses 0.4, once again in parentheses, and also to solve for X intercept. We need to note in order, once again, in order to find our X intersect, we need to recall and note that we need to set Y equal to 0 in our original equation. So when we do that, we will find that 0 is equal to X minus 4 divided by X plus 1. So when we rearrange this equation to isolate and solve for X all by itself, we will find that X is simply equal to 4. So that will mean that our X intercept is going to be 4.0. Fantastic. So, with that said, we now need to determine a few other coordinate points to plot on our graph to help us plot our curve. So, in order to solve for other points on our graph, let us choose different values for X. So, let us know that when X is equal to -2, That will mean that F of -2 is going to be equal to -2 minus 4 divided by -2 plus 1, which will be equal to 6. And in an effort to save time, I'm not going to keep writing out the equation all the way every single time, but let's say that X is equal to -6, that will mean that F of -6, so that means if you plug in a value of -6 in for X, that will mean that F of -6 will be simply equal to 2. So now, We can officially start to graph our function. So looking at our blank coordinate plane here as a reference, we need to note that for the asymptotes, we need to draw vertical, horizontal, and or slant asymptotes as dashed lines, be as a dotted or a dashed line indicates an asymptote, and that indicates to your professor or teacher that it's an asymptote. So that's very important. It has to be a dotted line, because if it's a solid line, that it could be mistaken as a curve. And also we need to recall and note that for the intercepts and other points, we need to plot them on our graph and then connect them with a smooth curve, making sure that they approach the asymptotes, but they never cross or touch our vertical asymptotes or horizontal asymptotes. So essentially they get very, very close to our asymptotes, but they do not touch or intersect or asymptotes. So, with that said, I've gone ahead and took our rational function for Y and plugged it into my graphing software of choice to produce the following graph. And as you can see, we have our asymptotes represented by green dotted or dashed lines for X equals -1 and Y equals 1. So Y equals 1 is our horizontal ascentote, and then our vertical ascentote is X equals -1. Then we have our curves, which are connected by a smooth curve between our points. We have -6.2 and 2.6 with our curve. Now as you can see, it gets really close to our asymptotes, but it doesn't touch or intersect them. We also have our points plotted at 0.4 and at 4.0. And once again, they're connected with a smooth curve, which is represented in red. So our curve for the rational function for Y equals X minus 4 divided by X + 1 is represented in red. And as you can see, once again, it gets really close to our asymptotes, but they do not touch or intersect them. That is very important when you sketch your graph that you don't accidentally cross your dotted line or go over because you will lose points as a result. And that's it. So once again, to quickly recap. So once again, to quickly recap, That means just to quickly go over everything that we've learned together, our final answers without a doubt, so we already determined our graph, but to quickly go over our horizontal class until once again is at Y equals 1, our vert class until it is at X equals -1. Our X intercept is at parentheses 4.0, and our Y intercept is at parentheses 0.4. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Graphing Simple Rational Functions

Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.

y = (x + 3)/(x + 2)

Key Concepts

Rational Functions

Asymptotes

Dominant Terms

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